Chapter 2, Problem 1RE

To determine

Solve the equation

  $\frac{3x+1}{5x+7}=\frac{6x+11}{10x-3}$

Answer to Problem 1RE

  $x=\frac{5}{6}$

Explanation of Solution

Given:

The given equation is

  $\frac{3x+1}{5x+7}=\frac{6x+11}{10x-3}$

Concept Used:

To simplify an equation of form:

  $\frac{P\left(x\right)}{Q\left(x\right)}=\frac{R\left(x\right)}{S\left(x\right)}$

Use cross multiplication, as

  $P\left(x\right)S\left(x\right)=R\left(x\right)Q\left(x\right)$

And solve it further to find value of variable x

Calculation:

To solve the equation

  $\frac{3x+1}{5x+7}=\frac{6x+11}{10x-3}$

First, simplify this equation by cross multiplying the denominators, as

  $\begin{array}{l}\left(3x+1\right)\left(10x-3\right)=\left(6x+11\right)\left(5x+7\right)\\ 30x^2+10x-9x-3=30x^2+42x+55x+77\\ 30x^2+x-3=30x^2+97x+77\\ \Rightarrow 30x^2+x-3-30x^2-97x-77=0\end{array}$

Now, this can be further simplified, as

  $\begin{array}{l}\Rightarrow \left(30-30\right)x^2+\left(1-97\right)x-\left(3+77\right)=0\text{ adding/subtacting terms of like power of }x\\ \Rightarrow 0x^2+96x-80=0\\ \Rightarrow 96x=80\\ \Rightarrow x=\frac{80}{96}\\ \Rightarrow x=\frac{20}{24}\\ \Rightarrow x=\frac{5}{6}\end{array}$

Hence, $x=\frac{5}{6}$ is the root of given equation.